What's the formula for this series for $\pi$?

Question

These continued fractions for $\pi$ were given here,

$$\small \pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}} = \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1} = \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1 $$

$$\small \pi = 3 + \cfrac{1^2} {6+\cfrac{3^2} {6+\cfrac{5^2} {6+\ddots}}} = 3 - \sum_{n=1}^\infty \frac{(-1)^n} {n (n+1) (2n+1)} = 3 + \frac{1}{1\cdot 2\cdot 3} - \frac{1}{2\cdot 3\cdot 5} + \frac{1}{3\cdot 4\cdot 7} - \cdots\tag2 $$

$$\small \pi = \cfrac{4} {1+\cfrac{1^2} {3+\cfrac{2^2} {5+\cfrac{3^2} {7+\ddots}}}} = 4 - 1 + \frac{1}{6} - \frac{1}{34} + \frac {16}{3145} - \frac{4}{4551} + \frac{1}{6601} - \frac{1}{38341} + \cdots\tag3$$

Unfortunately, the third one didn't include a closed-form for the series. (I tried the OEIS using the denominators, but no hits.)

Q. What's the series formula for $(3)$?

Answer 1

The third one should be obtained from $4.1.40$ in A&S p.68 using $z:=ix$ (from Euler I think not sure) :

$$-2\,i\,\log\frac{1+ix}{1-ix} = \cfrac{4x} {1+\cfrac{(1x)^2} {3+\cfrac{(2x)^2} {5+\cfrac{(3x)^2} {7+\ddots}}}} $$ Except that the expansion of the function at $x=1$ is simply your expansion for $(1)$.

Some neat variants :

$$\varphi(x):=\int_0^{\infty}\frac{e^{-t}}{x+t}dt= \cfrac{1} {x+1-\cfrac{1^2} {x+3-\cfrac{2^2} {x+5-\cfrac{3^2} {x+7-\ddots}}}}$$ $$\text{the previous one was better for large $x$...}$$ $$\int_0^{\infty}e^{-t}\left(1+\frac tn\right)^n\,dt=1+ \cfrac{n} {1+\cfrac{1(n-1)} {3+\cfrac{2(n-2)} {5+\cfrac{3(n-3)} {7+\ddots}}}}$$

$$\sum_{k=0}^\infty\frac 2{(x+2k+1)^2}= \cfrac{1} {x+\cfrac{1^4} {3x+\cfrac{2^4} {5x+\cfrac{3^4} {7x+\ddots}}}}$$ $$\text{and thus $\dfrac{\zeta(2)}2$ for $x=1$ (Stieltjes)}$$

$$\text{The last one was obtained after division by $n$ at the limit $n=0$ :}$$ $$\begin{align} \int_0^1\frac{t^{x-n}-t^{x+n}}{1-t^2}dx&=\sum_{k=0}^\infty\frac 1{x-n+2k+1}-\frac 1{x+n+2k+1}\\ &=\cfrac{n} {x+\cfrac{1^2(1^2-n^2)} {3x+\cfrac{2^2(2^2-n^2)} {5x+\cfrac{3^2(3^2-n^2)} {7x+\ddots}}}}\\ \end{align}$$

Your continued fraction appears too in a neat and recent book by Borwein, van der Poorten, Shallit, Zudilin "Neverending Fractions: An Introduction to Continued Fractions" at the end of the pages $167-169$ reproduced for convenience here (hoping there is no problem with that...) :

Answer 2

Given the symmetric continued fraction found in this post

$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}$$

Your continued fraction $(3)$ is a special case when $a=b=1$.

Moreover ,it has a beautiful q-analogue

$$\begin{aligned}\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}\end{aligned}$$ found here and here.